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For constant component masses, the center of mass frame has ${\mathbf{r}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=-(M_1/M)\mathbf{r}$. Newton's law of universal gravitation gives the relative equation $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Substitution in the component angular momenta yieldsHere $\mu =M_1{M}_2/M$ is the reduced mass, while $\mathbf{h}$ is specific angular momentum.

The printed total-energy expression lacks a factor of $\mu$ in its kinetic term. With the stated physical separation coordinate, the dimensionally consistent total two-body orbital energy isThe second expression is specific orbital energy; these two conventions must not be mixed. Its conservation follows directly:For the eccentricity vector, differentiate $GM\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}-GM\hat{\mathbf{r}}$. Since $\mathbf{h}$ is constant,The cancellation is the vector triple product identity. Thus all three corrected Kepler integrals are conserved. Dotting the eccentricity relation with $\mathbf{r}$ also gives the useful orbit identity

For ballistic streamline focusing by a moving star, work in the star's rest frame and take the incoming gas to have velocity $v{\mathbf{e}}_x$, with downstream behind the star at $x>0$. A streamline with impact parameter $d>0$ approaches from $(-\infty ,d,0)$. Its upstream specific angular momentum and eccentricity vector areAt its downstream axis crossing, $\mathbf{r}=b{\mathbf{e}}_x$. The orbit identity gives $b=d^2{v}^2/(GM)-b$, hence the collision distance isAt that point, $h_z=b{u}_y=-dv$, while the $y$ component of $GM\mathbf{e}=\mathbf{u}\times \mathbf{h}-GM\hat{\mathbf{r}}$ gives $dv{u}_x=d{v}^2$. ThereforeA symmetric streamline with opposite impact parameter has the opposite transverse velocity. Their shock wave removes the opposing transverse motion while preserving the common downstream component. Thus the remaining velocity is $v{\mathbf{e}}_x$ in the star frame. In the frame where the undisturbed medium is stationary, adding the star's velocity gives zero velocity immediately after this idealized collision.

After transverse kinetic energy is dissipated, the remaining specific orbital energy is ${\epsilon }_{\mathrm{after}}=v^2/2-GM/b$. It is negative if $b<2GM/v^2$, or equivalently if the upstream impact parameter satisfies $d<d_{\mathrm{cap}}=2GM/v^2$. This is the post-shock capture criterion in ballistic accretion; bound axial streams can return to the star. The captured incident mass flux through the corresponding disk gives the ballistic Bondi--Hoyle--Lyttleton accretion rate:This is the pressureless, dissipative capture model, with the shock's transverse energy unavailable to unbind the wake.

For wind mass transfer in a binary star, steady isotropic donor mass loss gives the local wind mass density at the companion:A fast wind has $v_w\gg 2\pi a/P$ and a capture scale small compared with $a$, so its relative incident speed is $v_w$ to leading order. Apply the same Bondi--Hoyle--Lyttleton accretion rate with accretor mass $M_2$ to obtainFinally, Kepler's third law eliminates the unquoted separation, $a=[G(M_1+M_2)P^2/(4{\pi }^2){]}^{1/3}$, giving the rate in terms of the stated orbital period:Equivalently, the fast-wind accretion fraction in a circular binary is $[M_2/(M_1+M_2){]}^2(v_{\mathrm{orb}}/v_w{)}^4$, with $v_{\mathrm{orb}}=2\pi a/P$ the relative circular orbital speed. Its smallness is consistent with using an almost undisturbed isotropic stellar wind.

Interpret projection from very far away as the asymptotic incoming state at $r\to \infty$. For $v>0$ and impact parameter $b>0$, conservation of energy and conservation of angular momentum give $E=v^2/2$ and $|h|=bv$. At the closest approach the radial velocity vanishes. With $k=-\kappa <0$, the central-force radial turning point equation isOnly the positive root is physical. Put $q=\kappa /v^2$. The pericentre distance and the purely tangential speed there areThus $p<b$ and $u>v$: attraction bends the path inward and increases the speed. With positive energy, the Kepler orbit is a hyperbola. Measured from the pericentre direction, its polar coordinates satisfy $r=h^2/[\kappa (1+e\cos \varphi )]$ with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$.

The hyperbola shown has a nonzero impact parameter. For $b=0$, the attractive radial orbit instead reaches the singular origin; there is no regular turning point with a finite closest-approach speed.

Use an edge-on orbit and an equatorial, effectively central chord. Near the foreground crossing the radial velocity is along the line of sight; the projected transverse speed is $v_{\perp }=h/r$. In the small stellar-angular-radius limit, $r\gg R_{\star }$, the crossing time is $2R_{\star }/v_{\perp }=2R_{\star }r/h$.

The cross-sectional-area current therefore places an areain front of the stellar disk. For an optically thin wire and a uniformly bright stellar disk, transit dimming by an optically thin orbital wire givesThis is a linear occultation estimate, requiring negligible overlap and a fractional dimming much smaller than one. A finite impact parameter shortens the chord; limb darkening, finite wire thickness and variation of $r$ across a large stellar angular extent modify the coefficient. The estimate cannot be extrapolated to dimming greater than unity.

Work in the planet's rest frame, neglecting its recoil and tidal forces during the close encounter. A hyperbolic Kepler orbit has positive specific orbital energy $\epsilon =v_{\infty }^2/2$ and specific angular momentum $h=b{v}_{\infty }$, where $b$ is the impact parameter. Its orbital eccentricity isAt the pericentre distance, $r_{\min }=h^2/[\mu (1+e_h)]$. Substituting $h^2={\mu }^2(e_h^2-1)/v_{\infty }^2$ yieldsThe two asymptotes have true anomalies $\pm f_{\infty }$, where $\cos f_{\infty }=-1/e_h$. Their directions of motion give the gravitational scattering angle $\theta =2f_{\infty }-\pi$, so $\sin (\theta /2)=-\cos f_{\infty }=1/e_h$. ThereforeAs a useful independent form, $b^2=r_{\min }^2+2\mu r_{\min }/v_{\infty }^2$, displaying gravitational focusing. Here $\mu =G{M}_p$ to the stated small-comet accuracy; the exact relative-coordinate formula replaces it by $G(M_p+M)$.

The stellar relaxation time is the time on which accumulated discrete gravitational encounters change a typical star's velocity by an amount comparable to its original velocity. It concerns two-body relaxation, rather than the much faster orbital evolution in a smooth gravitational potential. A collisionless stellar system requires this time to greatly exceed the duration being studied.

In the straight-line impulse approximation, an encounter at relative speed $u$ has transverse acceleration $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. ThusA small-angle gravitational encounter needs $\Delta v_{\perp }\ll u$. The transition to order-one deflections is therefore $b\sim Gm/u^2$, ignoring factors such as the equal-mass relative deflection. For the spherical estimate $u\sim v$,This is the lower cutoff of the weak-scattering estimate; closer encounters actually occur and require strong-scattering treatment.

The number of encounters in time $T$ with impact parameters in $[b,b+db]$ is $2\pi bdbnuT$. Independent random transverse directions make mean kicks cancel while their variances add. ConsequentlyEach logarithmic interval contributes equally, giving the Coulomb logarithm in stellar dynamics. Defining a deflection time by variance $v^2$ and setting $u=v$ gives $v^3/(8\pi G^2{m}^{2}n\log \Lambda )$. The paper instead uses an order-one normalization three quarters of this estimate:Both have the same physical scaling. The specified characteristic speed, the word “comparable”, the velocity-distribution average and strong-encounter cutoff do not fix that numerical coefficient uniquely. The crude equal-speed impulse calculation must not be claimed to determine $3/(32\pi )$ exactly.

For a self-gravitating, approximately virialized system, the virial theorem gives $v^2\sim GNm/R$. With $n=3N/(4\pi R^3)$, $\Lambda \sim N$ and the stellar crossing time $t_{\mathrm{dyn}}=R/v$, substitution into the paper's convention givesAn externally dominated gravitational potential or a different structural constant changes this substitution. For an N-body simulation lasting $T=K{t}_{\mathrm{dyn}}$, a tolerable fractional velocity-squared diffusion $\epsilon \ll 1$ requires $T/t_{\mathrm{rel}}\lesssim \epsilon$, henceThere is no unique smallest $N$ without $K$, the error tolerance and the force prescription. For scale, equality at $t_{\mathrm{rel}}/t_{\mathrm{dyn}}=100$ occurs near $N=7.1\times 10^3$; $N=10^5$ gives about $1.1\times 10^3$ crossing times. A calculation lasting 100 crossing times therefore needs substantially more than the first threshold to have negligible relaxation. Gravitational softening can raise the effective cutoff and reduce artificial scattering, but it also sets the spatial force resolution.

Put $u=\lambda v$ and let $n$ now mean number per area. During time $T$, the two strips of impact parameters between $b$ and $b+db$, on opposite sides of the trajectory, have combined area $2uTdb$. ThusThe same straight-line impulse approximation gives a kick $2Gm/(bu)$. Summing its squared size over these encounters givesassuming $b_{\max }\gg b_{\min }$. The relevant change is now the small random speed $u$, not the full circular speed $v$. Setting $D_v{t}_{\mathrm{rel}}\sim u^2$ yieldsThe razor-thin disk relaxation integral is dominated by nearby encounters, rather than equally by logarithmic intervals. Because kicks cease to be weak near the cutoff, its numerical coefficient is an estimate.

Here the weak-deflection cutoff must use the relative speed: $b_{\min }\sim Gm/u^2=Gm/({\lambda }^2{v}^2)$, which is larger than the spherical cutoff. Thus $t_{\mathrm{rel}}\sim u/(Gmn)$. Using $n\sim N/R^2$ and $v^2\sim GNm/R$ from the virial theorem givesFor example $n=N/(\pi R^2)$ gives the order-one coefficient $\pi /8$ in this last substitution; the paper's approximate equality suppresses such structural factors. The cancellation of $N$ is the central result.

An unsoftened point-particle system confined to an infinitesimally thin plane is therefore not collisionless over orbital times for $\lambda \sim 0.1$ in this encounter model. At fixed total mass, increasing $N$ decreases $m$ and the cutoff together; the enhanced importance of close planar encounters cancels the usual benefit of increasing particle number. This is a singular zero-thickness limit. A fixed nonzero gravitational softening length or physical thickness restores a different $N$ dependence, so the conclusion is not a prohibition on collisionless numerical models of disks.

With negligible comet radius and mass, the planet's radius and escape velocity areGravitational focusing changes the limiting collision impact parameter from $R_p$ to $R_p\sqrt{1+v_{\mathrm{esc},\mathrm{p}}^2/v_{\mathrm{rel}}^2}$. Thus it is negligible when $v_{\mathrm{esc},\mathrm{p}}^2\ll v_{\mathrm{rel}}^2$.

Put $s=q/a_p$ and $\gamma =\sqrt{3-2\sqrt{2s}}$ in the $Q\gg a_p$ prograde approximation. Substituting $v_{\mathrm{rel}}=\gamma \sqrt{G{M}_{\star }/a_p}$ givesEquivalently $2(M_p/M_{\star })(a_p/R_p)\ll {\gamma }^2$. For $q\ll a_p$, use ${\gamma }^2\simeq 3$. If negligible gravitational focusing is required for every coplanar near-parabolic orbit, including nearly tangential prograde encounters, use the smaller value ${\gamma }^2=(\sqrt{2}-1{)}^2$.

In a planetary gravitational assist, the incoming and outgoing relative speeds have the same magnitude. If $\mathbf{u}$ is the planet-frame velocity, the heliocentric specific orbital energy change is consequentlynot an arbitrary addition of $|\Delta \mathbf{u}|$ to the heliocentric speed. The initial binding energy per unit mass is $G{M}_{\star }/(Q+q)\simeq G{M}_{\star }/Q$.

For a weak deflection, $b\gg b_{90}=G{M}_p/v_{\infty }^2$, the hyperbolic scattering formula gives $|\Delta \mathbf{u}|\simeq 2G{M}_p/(b{v}_{\infty })$. Its direction is perpendicular to the incoming relative velocity. At the planet, in the near-parabolic prograde limit,On the favorable side of the planet the energy gain is approximately $2G{M}_p{v}_p\kappa /(b{v}_{\infty })$. Equating this to the binding energy gives the ejection impact parameterOnly one sign of impact parameter gains energy for each radial branch. Noncolliding ejections have $R_p<|b|<b_{\mathrm{ej}}$, giving width $(b_{\mathrm{ej}}-R_p{)}_{+}$, where $x_{+}=\max (x,0)$. For $b_{\mathrm{ej}}\gg R_p$, the planetary ejection of a comet rate is approximately ${\Gamma }_{\mathrm{ej}}={\Sigma }_c{v}_{\infty }{b}_{\mathrm{ej}}$. With ${\mu }_p=M_p/M_{\star }$ this givesThe scaling applies when $b_{90}\ll b_{\mathrm{ej}}\ll r_H$ so that the local two-body encounter and weak-deflection approximations are consistent. The favorable-side coefficient assumes a coplanar orbit; averaging over other encounter geometries changes it. Cumulative diffusion by many subthreshold encounters is not included in this one-encounter estimate.

Ignoring gravitational focusing, a galaxy sweeps a cylinder of volume $\pi R^2{v}_{\mathrm{rel}}T$. Multiplying this by the number density $N$ of target galaxies and averaging the relative speed gives the mean encounter countFor independent encounters modeled by a Poisson process, the probability of at least one galaxy merger is $1-e^{-\mu }$. The printed expression $P=\mu$ is its rare-encounter approximation $\mu \ll 1$; it is not an exact probability for arbitrary $T$.

For an illustrative present-day field population take an effective merger impact parameter $R=20\mathrm{kpc}=0.02\mathrm{Mpc}$, $N=10^{-2}{\mathrm{Mpc}}^{-3}$, and $⟨v_{\mathrm{rel}}⟩=200\mathrm{km}{\mathrm{s}}^{-1}$. These are assumed order-of-magnitude inputs, not precise observational measurements. Taking a Hubble time of $T\simeq 1.4\times 10^{10}\mathrm{yr}\simeq 4.4\times 10^{17}\mathrm{s}$, the given distance conversion yieldsand henceThus the geometric field estimate is of order $10^{-5}$ to $10^{-4}$ per Hubble time. It scales as $R^{2}N⟨v_{\mathrm{rel}}⟩T$ and is very sensitive to environment and the adopted effective merger radius. Enhanced density in groups, gravitational focusing, and the evolution of the galaxy population are all omitted; this number is not a prediction of the full cosmological merger fraction.

Now use the stipulated rapid, distant encounter. The consistent rectilinear trajectory isWith the stated impact parameter along $x$ and velocity along $z$, this trajectory lies in the $xz$-plane. The original PDF's reference to the $xy$-plane is a typo; the local TeX also corrupts the impact-parameter direction. The original PDF fixes that direction as $x$.

Use the positive potential $G{m}_p/|\mathbf{R}-\mathbf{r}|$, whose gradient is the attractive acceleration under the question's sign convention. Its multipole expansion isThe first term has no force; the linear term accelerates the entire galaxy and disappears in the frame following its centre of mass. The leading internal tidal potential is thereforeThis is a quadrupole approximation valid for $r\ll p$, rather than an exact equality for every $r<p$. Its acceleration tidal tensor acts on $\mathbf{r}$ asIn the impulse approximation, each stellar position is held fixed during the flyby and the velocity kick is the integral of this acceleration. Put $s=vt$; terms odd in $s$ integrate to zero, whileThe integrated diagonal coefficients of the tidal tensor are consequently $(2,-2,0)/(v{p}^2)$, givingThe encounter stretches the galaxy along the impact parameter, compresses it along $y$, and gives no net leading kick along the flyby direction.

Denote the pre-encounter stellar velocity by $\mathbf{u}$, to distinguish it from the relative flyby speed $v$. The instantaneous change of specific kinetic energy isUncorrelated kicks with zero mean cross term imply $⟨\mathbf{u}\cdot \Delta \mathbf{v}⟩=0$, so the phase-averaged heating at a specified position isFor an individual star the cross term need not vanish: the formula is an ensemble or orbital-phase average. Integrating over a spherical galaxy of mass $m_g$, symmetry gives $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=⟨r^2⟩/3$. Therefore the total tidal heating isFor two identical galaxies, each receives this heating with $m_p=m_g$. Adding both contributions gives

In the centre of mass frame, equal masses approach with speeds $v/2$ if $v$ is their relative speed at infinity. The initial orbital energy iswhere $m_g/2$ is the reduced mass. The orbital potential energy vanishes at infinite separation. Within the weak-deflection approximation this asymptotic speed is also the nearly constant speed used in the flyby calculation.

Conservation of energy transfers the positive internal heating out of the relative orbit. Tidal capture of galaxies occurs in this model if $E_{\mathrm{orb}}-\Delta E_{\mathrm{int}}<0$, which givesThe equality is marginal capture. A bound pair still needs subsequent evolution to coalesce, so the result is an approximate capture criterion used here as a merger criterion. Its right-hand side has dimensions of length times speed, as required.

The impulse approximation requires the flyby duration $\tau \sim p/v$ to be short compared with the stellar dynamical time ${\Omega }^{-1}$, namely $v/p\gg \Omega$. If $v/p<\Omega$, a star moves substantially during the encounter. Its orbital adiabatic invariance of an orbital action suppresses net heating by a slowly varying tidal field: adiabatic shielding of tidal encounters replaces the frozen-position calculation. The divergent $v^{-2}$ heating predicted by extrapolating the impulsive formula to slow encounters is therefore spurious. Close passages with $p$ comparable to the galaxy size or strong gravitational focusing also lie outside the derivation.

In the classical experiment, particles whose impact parameter lies between $b$ and $b+db$ occupy an incident annulus of areaFor a spherically symmetric potential, azimuthal symmetry scatters this annulus into the solid-angle stripThus the central-potential classical differential cross-section isIf the relation between $b$ and $\theta$ has several branches, one sums this expression over every branch. The corresponding diagram consists of an incoming line offset by $b$ from the centre, an outgoing asymptote making angle $\theta$ with the incident direction, and the annulus and solid-angle strip just described.

For quantum scattering by a central potential, the far-field wavefunction iswhere $f(\theta )$ is the scattering amplitude. The incident plane wave has probability currentTo leading order in $1/r$, the scattered spherical wave has radial currentThe scattered rate through area $r^{2}d\Omega$ divided by the incident flux is therefore the quantum differential cross-section from a scattering amplitude:

In the planetocentric hyperbolic Kepler orbit, the speed at infinity is $\Delta v_{pc}$ and the impact parameter is $b$. Conservation of specific angular momentum givesConservation of specific orbital energy at the moon's radius giveswhere $v_m^2=G{M}_p/a_m$. ThusThe moon moves tangentially at $v_m$, soTherefore the correctly dimensioned reading of the displayed result is

The radial first integral for an equatorial null geodesic in Schwarzschild spacetime isSince $L=r^2\dot{\varphi }$, regarding $r$ as a function of $\varphi$ givesSet $y=1/r$. Then $dr/d\varphi =-y^{\prime }/y^2$ and therefore $\dot{r}=-L{y}^{\prime }$. Division by $L^2$ gives the equatorial null orbit in Schwarzschild spacetimewhere $b=L/E$ is the impact parameter. Differentiating with respect to $\varphi$ givesBy continuity this yields

Chandrasekhar dynamical friction is the drag exerted by the overdense gravitational wake that a massive body creates in a background of lighter particles. It transfers orbital energy and angular momentum to the host, causing satellites, star clusters, and massive black holes to spiral inward and promoting galaxy mergers. In a homogeneous isotropic Maxwellian background its force iswhere $\log \Lambda$ is the Coulomb logarithm in stellar dynamics. Its scaling can be reconstructed from the strong-deflection impact parameter $b_{\mathrm{cr}}\sim GM/v^2$: the encountered mass rate is $\sim \rho v\pi b_{\mathrm{cr}}^2$, and multiplying by momentum change $\sim v$ gives $F_{\mathrm{df}}\sim G^2{M}^2\rho /v^2$.

For a spherical host with a flat galaxy rotation curve,This is a singular isothermal sphere with one-dimensional dispersion $\sigma =v_c/\sqrt{2}$, so $X=1$. DefineFor a constant-mass satellite on a circular orbit, the drag magnitude and its torque areIt follows thatThe quadratic radius dependence and inverse mass dependence explain why massive nearby satellites merge much faster than light or distant ones.

Let the satellite also have a flat internal rotation curve of speed $v_s$. Equating its edge density to the host density gives the tidal radiusBecause $M_s(<R_t)=v_s^2{R}_t/G$, the bound mass decreases linearly:Under the question's literal closure that the derivative of the remaining satellite's total orbital angular momentum equals the frictional torque,and henceIf stripped material is explicitly assigned the satellite's instantaneous specific orbital angular momentum, the balance for the bound remnant is instead $M_{s}d(v_{c}r)/dt=-r{F}_{\mathrm{df}}$; that convention gives $t_{\mathrm{df},\mathrm{strip}}=2t_{\mathrm{df},0}$. Both treatments show the robust point: tidal stripping weakens the drag as the orbit shrinks and substantially delays coalescence. In less idealized profiles the mass can fall faster than linearly, producing dynamical-friction stalling by tidal stripping.

Let $v_K=\sqrt{G{M}_{\ast }/a_p}$. At periapsis the nearly parabolic planetesimal and planet have parallel speeds $\sqrt{2}{v}_K$ and $v_K$, soFor a passage just outside the planet, the impact parameter is $b\simeq \delta a_p$. In the small-angle limit of the supplied gravitational scattering angle,Thus

During a weak encounter, approximate one star's trajectory by a straight line with speed $v$ and impact parameter $b$. At longitudinal coordinate $z=vt$, the transverse acceleration isIntegrating from $t=-\infty$ to $\infty$ givesHere $Gm/b^2$ sets the gravitational acceleration, $b/v$ is the encounter duration, and the geometric projection produces the displayed transverse impulse.

In one crossing, the number of encounters with impact parameters in $(b,b+db)$ is, up to the system-geometry convention,Uncorrelated impulses add in mean square, soThe virial theorem gives $v^2\sim GNm/R$. Taking $b_{\max }\sim R$ and the strong-deflection scale $b_{\min }\sim Gm/v^2\sim R/N$ gives the Coulomb logarithm in stellar dynamics $\log \Lambda \sim \log N$ andVelocity memory is lost when the cumulative change reaches $v^2$, afterConsequently the two-body relaxation time iswith an order-unity prefactor depending on density profile and convention. The system is a collisional stellar system when $t_{\mathrm{rel}}$ is shorter than its age or the evolutionary timescale being studied.

A gravitational assist ejects a weakly bound comet when the increase in its specific orbital energy exceeds its binding energy. For near-parabolic coplanar encounters, the weak-deflection ejection impact parameter scales as $(M_p/M_{\star })Q$. Combining this with the phase-mixed comet surface density gives a one-encounter ejection time proportional to $Q^{1/2}$, within the local two-body encounter approximation.

For a particle initially having semimajor axis one and encountering a circular planet at radius one, the incoming stellar-frame speed is one. A conservative encounter at planet-relative speed $u$ changes the stellar-frame velocity by at most $2u$. Since stellar escape requires speed $\sqrt{2}$, a necessary bound is $u>(\sqrt{2}-1)/2$. Retaining the outgoing relative-velocity geometry sharpens the necessary bound to $u>\sqrt{2}-1$. Neither inequality by itself guarantees a realizable deflection at a specified impact parameter.